CEE 345 - Benjamin - Winter 2012 - Homework 2 Solutions(1)

# CEE 345 - Benjamin - Winter 2012 - Homework 2 Solutions(1)...

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2 2 2 22 3 2 44 55 2 2 ss 1 f t 1 m i n f t - m i n f t 73.4 73.4 3.64x10 3.64x10 ft ft 7.48 gal 60 s gal gpm ±± §· § · ¨¸ ¨ ¸ ©¹ © ¹ 42 2 ft 50 ft 3.64x10 gpm pump hQ ± ² The equation of the pump curve, with TDH in ft and Q in gpm is: 2 ft TDH 180 ft 6.10x10 gpm Q ± ± The operating point occurs at the flow rate where h pump = TDH, so ft ft 50 ft 3.64x10 180 ft 6.10x10 gpm gpm QQ ² ± 365 gpm Q 3. The three data points characterizing the pump curve can be plotted as shown in the figure below, and a smooth curve can be drawn through them. We are told that when the static head is 15 m, the flow rate is 150 L/s, and the pump curve indicates that when the flow rate is 150 L/s, the pump adds 15.7 m of head to the water. Therefore, the headloss through the pipe under these conditions (equal to the sum of the static head and the head added by the pump) is 30.7 m. Because we are told that the system curve is
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CEE 345 - Benjamin - Winter 2012 - Homework 2 Solutions(1)...

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